Maximum value of a function

This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 231
Joined: Thu Apr 12, 2007 2:45 am
Thanked: 5 times
Followed by:1 members

Maximum value of a function

by winnerhere » Sun Jul 04, 2010 5:18 am
Y = 25x - x^2

Find the value of X , for which Y attains maximum value.

How to proceed with this question

Thanks in Advance,
Sai
Last edited by winnerhere on Sun Jul 04, 2010 5:41 am, edited 1 time in total.

Senior | Next Rank: 100 Posts
Posts: 95
Joined: Tue Jun 08, 2010 11:27 am
Thanked: 7 times

by singhpreet1 » Sun Jul 04, 2010 5:34 am
winnerhere wrote:Y = x^2 - 25x

Find the value of X , for which Y attains maximum value.
Sai: if we take the value of x as negative we can get the maximum value of Y, though what are the answer choices?

Please post the same along with OA.

Preet

User avatar
Legendary Member
Posts: 1893
Joined: Sun May 30, 2010 11:48 pm
Thanked: 215 times
Followed by:7 members

by kvcpk » Sun Jul 04, 2010 5:38 am
winnerhere wrote:Y = x^2 - 25x

Find the value of X , for which Y attains maximum value.

How to proceed with this question

Thanks in Advance,
Sai
Y = x(x-25)
when x is less than -25 I beleive, Y will Attain its maximum and will keep increasing..

Are you sure about the question? What is the source!!

Master | Next Rank: 500 Posts
Posts: 231
Joined: Thu Apr 12, 2007 2:45 am
Thanked: 5 times
Followed by:1 members

by winnerhere » Sun Jul 04, 2010 5:39 am
Preet,

The answer choices are,

a) 14.5

b) 14

c) 13

d) 12.5

OA : d

Master | Next Rank: 500 Posts
Posts: 231
Joined: Thu Apr 12, 2007 2:45 am
Thanked: 5 times
Followed by:1 members

by winnerhere » Sun Jul 04, 2010 5:41 am
kvcpk and preet,

oops sorry,

its y = 25x - x^2

User avatar
Legendary Member
Posts: 1893
Joined: Sun May 30, 2010 11:48 pm
Thanked: 215 times
Followed by:7 members

by kvcpk » Sun Jul 04, 2010 6:28 am
winnerhere wrote:kvcpk and preet,

oops sorry,

its y = 25x - x^2
Ok, in this case, Y will attain a maximum value if x(25-x) os maximum.
let p = (25-x)
then x+p = 25
given their sum, the product of two number is maximum, when they are equal..
(I did not put this rule in the exact terms.. I dont remember the exact terminology.. I request some experts to restate it as per textbooks.)

So product of x and p is maximum when x=p
So, 2x=25 -> x= 12.5

Senior | Next Rank: 100 Posts
Posts: 95
Joined: Tue Jun 08, 2010 11:27 am
Thanked: 7 times

by singhpreet1 » Sun Jul 04, 2010 6:49 am
winnerhere wrote:kvcpk and preet,

oops sorry,

its y = 25x - x^2
Personally,

i wouldnt even calculate anything for this one, the answer stares in the face for me out of the options: least value of positive X will help Y be maximum.

out of the options given. 12.5 is the lowest.

eg: assume x=10

y= 25*10- 10^2= 150

assume x=20

y= 25*20-20^2= 100

assume y=30

25*30-30^2= -150

therefore the sequence clearly shows that the value of Y is decreasing as X is increasing.

thats my personal way..though please use what suits you.

Preet

User avatar
GMAT Instructor
Posts: 1052
Joined: Fri May 21, 2010 1:30 am
Thanked: 335 times
Followed by:98 members

by Patrick_GMATFix » Sun Jul 04, 2010 7:10 am
winnerhere wrote:Y = 25x - x^2

Find the value of X , for which Y attains maximum value.
12.5. I'll explain below.
singhpreet1 wrote:i wouldnt even calculate anything for this one, the answer stares in the face for me out of the options: least value of positive X will help Y be maximum.

out of the options given. 12.5 is the lowest.
You are making a dangerous assumption that the least positive X will result in the maximum Y. This is not accurate. If X=1 then Y will not be greater than if X is 10.

Here is how I solved the question.

This is a quadratic function, meaning that the graph is a parabola. Furthermore, because the coefficient (the number in front) of X is negative (-1) the parabola opens down. The curve is shaped like a rainbow.

Parabolas are symmetrical around their "middle". The maximum value will be the value of X that is exactly halfway between the two points at which the parabola touches the horizontal x-axis (these are called the roots or solutions of the equation; the points where y=0). So let's find those points.

Y=25x - x^2. To solve any quadratic equation (to find its roots) you factor it ---> y = x (25-x). The solutions are the values of x that will make the equation 0. They are x=0 and x=25. This means that the parabola goes up, crossing the x-axis at 0, then reaches its maximum and comes back down, crossing the x-axis again at 25. The maximum is half-way between the two. The maximum value of Y is when X=25/2 --> 12.5

Note: although knowing how to handle parabolas in the xy-axis system can be useful on a few GMAT problems, this skill is not required for the GMAT and may not be worth investing a lot of time to learn if you have weaknesses in GMAT tested topics.

Hope that made sense,
-Patrick
  • Ask me about tutoring.

Master | Next Rank: 500 Posts
Posts: 265
Joined: Mon Dec 28, 2009 9:45 pm
Thanked: 26 times
Followed by:2 members
GMAT Score:760

by mj78ind » Sun Jul 04, 2010 7:44 am
Given the 2 mins timeline,

Approach 1
engineers have an advantage. Calculate the first differential wrt x of the quadratic, and that is your answer since the second differential is negative. First differential of the equation is 25/2 = 12.5

Approach 2
Plug in the numbers and you will see that 12.5 gives the maxima.

User avatar
Master | Next Rank: 500 Posts
Posts: 294
Joined: Wed May 05, 2010 4:01 am
Location: india
Thanked: 57 times

by amising6 » Sun Jul 04, 2010 7:49 am
winnerhere wrote:Y = 25x - x^2

Find the value of X , for which Y attains maximum value.

How to proceed with this question

Thanks in Advance,
Sai
Y = 25x - x^2
y=-(x^2-25x)

y=-(x^2-2*x*25/2)
y=-(x^2-(2*x*25/2 )+(25/2)^2 - (25/2)^2) ( (adding (25/2)^2) and subtracting (25/2)^2)))
y=-((x+25/2)^2-(25/2)^2) (a^2+b^2+2ab)
y=(25/2)^2)=625/4 max values
since square of a number is always gretaer than or equal to zero
(x+25/2)^2>=0
so x=-25/2
Ideation without execution is delusion

User avatar
Legendary Member
Posts: 1893
Joined: Sun May 30, 2010 11:48 pm
Thanked: 215 times
Followed by:7 members

by kvcpk » Sun Jul 04, 2010 8:13 am
kvcpk wrote:given their sum, the product of two number is maximum, when they are equal..
(I did not put this rule in the exact terms.. I dont remember the exact terminology.. I request some experts to restate it as per textbooks.)
Can any expert restate this for me in exact terms? Thank you!!

User avatar
Master | Next Rank: 500 Posts
Posts: 294
Joined: Wed May 05, 2010 4:01 am
Location: india
Thanked: 57 times

by amising6 » Sun Jul 04, 2010 8:31 am
kvcpk wrote:
kvcpk wrote:given their sum, the product of two number is maximum, when they are equal..
(I did not put this rule in the exact terms.. I dont remember the exact terminology.. I request some experts to restate it as per textbooks.)


Can any expert restate this for me in exact terms? Thank you!!
u r rite kvcpk

it says if sum of gven number is contant then therir product will be maximum if numbers as close as possible

if you have a+b=12
then ab i.e product will be maximum when and b are equal or close to each other
you can check if a=b=6
ab=36 i.e maximu m
next if a=7 b=5 ab=35 next maximum so on and so far
Ideation without execution is delusion

Legendary Member
Posts: 2326
Joined: Mon Jul 28, 2008 3:54 am
Thanked: 173 times
Followed by:2 members
GMAT Score:710

by gmatmachoman » Sun Jul 04, 2010 8:45 am
kvcpk wrote:
kvcpk wrote:given their sum, the product of two number is maximum, when they are equal..
(I did not put this rule in the exact terms.. I dont remember the exact terminology.. I request some experts to restate it as per textbooks.)
Can any expert restate this for me in exact terms? Thank you!!
Praveen bhai,
Probably this one should help u!!

Y = 25x - x^2

step 1 :find dy/dx

dy/dx = 25 - 2x

step 2: set dy/dx= 0

25 = 2x

x= 25/2 = 12.5

For max values d^2y/dx^2 (second degree differential) <0

For minumim values d^2y/dx^2 >0

So cross check by putting value of x in d^2y/dx^2

d^2y/dx^2 = -2 which is anyway less than zero.

Pick D.

On a side note dont worry GMAC never tests maxima & minima...

User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

by Stuart@KaplanGMAT » Sun Jul 04, 2010 9:14 am
kvcpk wrote:
kvcpk wrote:given their sum, the product of two number is maximum, when they are equal..
(I did not put this rule in the exact terms.. I dont remember the exact terminology.. I request some experts to restate it as per textbooks.)
Can any expert restate this for me in exact terms? Thank you!!
Hi!

This rule is easy to remember if you think of rectangles and squares.

For rectangles, given a fixed perimeter, you maximize the area of the shape by making it a square.

In algebraic terms, given a fixed value x, the maximum product that can be derived by two numbers that sum to x is:

(x/2)^2 = (x^2)/4.
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course

User avatar
Legendary Member
Posts: 1893
Joined: Sun May 30, 2010 11:48 pm
Thanked: 215 times
Followed by:7 members

by kvcpk » Sun Jul 04, 2010 9:56 am
Thanks for all your replies guys..

I wanted to confirm the rule that I was using.. I am answered now..

Thank you!!